Question

For real \(x\) and if \(x+\frac{1}{x}=2\cos\theta\), then \(\cos\theta\) is

• A. \(\pm 1\)
• B. \(\frac{1}{2}\)
• C. \(1\)
• D. \(\pm \frac{1}{2}\)

Hint:

For a quadratic equation to have real roots, always use discriminant \(D\geq 0\).

Answer 1

The Correct Option is A

Step 1: Given, \[ x+\frac{1}{x}=2\cos\theta \]

Step 2: Multiply by \(x\): \[ x^2+1=2x\cos\theta \] \[ x^2-2x\cos\theta+1=0 \]

Step 3: Since \(x\) is real, discriminant must be non-negative: \[ D=(-2\cos\theta)^2-4(1)(1)\geq 0 \] \[ 4\cos^2\theta-4\geq 0 \] \[ \cos^2\theta\geq 1 \]

Step 4: But for any angle, \[ -1\leq \cos\theta \leq 1 \] So, \[ \cos^2\theta=1 \] \[ \cos\theta=\pm 1 \] \[ \boxed{\pm 1} \]